Problem: Simplify and expand the following expression: $ \dfrac{5}{4k + 8}+ \dfrac{1}{3k - 24}+ \dfrac{3k}{k^2 - 6k - 16} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{5}{4k + 8} = \dfrac{5}{4(k + 2)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{1}{3k - 24} = \dfrac{1}{3(k - 8)}$ We can factor the quadratic in the third term: $ \dfrac{3k}{k^2 - 6k - 16} = \dfrac{3k}{(k + 2)(k - 8)}$ Now we have: $ \dfrac{5}{4(k + 2)}+ \dfrac{1}{3(k - 8)}+ \dfrac{3k}{(k + 2)(k - 8)} $ The least common multiple of the denominators is: $ 12(k + 2)(k - 8)$ In order to get the first term over $12(k + 2)(k - 8)$ , multiply by $\dfrac{3(k - 8)}{3(k - 8)}$ $ \dfrac{5}{4(k + 2)} \times \dfrac{3(k - 8)}{3(k - 8)} = \dfrac{15(k - 8)}{12(k + 2)(k - 8)} $ In order to get the second term over $12(k + 2)(k - 8)$ , multiply by $\dfrac{4(k + 2)}{4(k + 2)}$ $ \dfrac{1}{3(k - 8)} \times \dfrac{4(k + 2)}{4(k + 2)} = \dfrac{4(k + 2)}{12(k + 2)(k - 8)} $ In order to get the third term over $12(k + 2)(k - 8)$ , multiply by $\dfrac{12}{12}$ $ \dfrac{3k}{(k + 2)(k - 8)} \times \dfrac{12}{12} = \dfrac{36k}{12(k + 2)(k - 8)} $ Now we have: $ \dfrac{15(k - 8)}{12(k + 2)(k - 8)} + \dfrac{4(k + 2)}{12(k + 2)(k - 8)} + \dfrac{36k}{12(k + 2)(k - 8)} $ $ = \dfrac{ 15(k - 8) + 4(k + 2) + 36k} {12(k + 2)(k - 8)} $ Expand: $ = \dfrac{15k - 120 + 4k + 8 + 36k}{12k^2 - 72k - 192} $ $ = \dfrac{55k - 112}{12k^2 - 72k - 192}$